Question

lim of (sin x-1) / (x - pi/2) as x approaches Pi/2

Answer 1

\[\lim_{x \rightarrow \frac{\pi}{2}}\frac{\sin(x)-1}{x-\frac{\pi}{2}}\]
recall sin(pi/2)=1\[\lim_{x \rightarrow \frac{\pi}{2}}\frac{\sin(x)-\sin(\frac{\pi}{2})}{x-\frac{\pi}{2}}\]
does this look familiar?

Answer 2

Yes.

Answer 3

\[\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(x)|_{x=a}\]

Answer 4

so if f(x)=sin(x) then f'(x)=?

Answer 5

cos x

Answer 6

and what is cos(x) evaluated at x=pi/2

Answer 7

right?

Answer 8

yes f'(x)=cos(x)
but what is f'(a) where a=pi/2

Answer 9

0

Answer 10

yep :)

Answer 11

\[\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(x)|_{x=a} \\ \lim_{x \rightarrow \frac{\pi}{2}}\frac{\sin(x)-\sin(\frac{\pi}{2})}{x-\frac{\pi}{2}}=\cos(x)|_{x=\frac{\pi}{2}}=\cos(\frac{\pi}{2})=0 \]

Answer 12

and I could have wrote it as f'(a) but for some reason that confuses people

Answer 13

and if you knew l'hosptal you could have done that too

Answer 14

f'(a) is a bit of an odd notation. Your version is much more clear.

f(a) is a constant. The 1st derivative of f(a) should be zero.

Answer 15

f'(a) doesn't mean f itself is a constant function
f'(a) means to find the general form for f'(x) and evaluate x at x=a